Problem: Let $f$ be a vector-valued function defined by $f(t)=\left(-4\sqrt{t},-\dfrac3t\right)$. Find $f'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\dfrac2{\sqrt t},\dfrac3{t^2}\right)$ (Choice B) B $-4\sqrt {t}+\dfrac3{t^2}$ (Choice C) C $\left(4t,3\ln(t)\right)$ (Choice D) D $\left(4\sqrt {t},-\dfrac3{t^2}\right)$
Explanation: $f$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $f(t)=\left(-4\sqrt{t},-\dfrac3x\right)$. Let's differentiate the first expression: $\begin{aligned}\dfrac{d}{dt}(-4\sqrt{t})&=-\dfrac12\cdot\dfrac4{\sqrt t} \\\\&=-\dfrac2{\sqrt t}\end{aligned}$ Let's differentiate the second expression: $\dfrac{d}{dt}\left(-\dfrac3t\right)=\dfrac3{t^2}$ Now let's put everything together: $\begin{aligned} f'(t)&=\left(\dfrac{d}{dt}(-4\sqrt{t}),\dfrac{d}{dt}\left(-\dfrac3t\right)\right) \\\\ &=\left(-\dfrac2{\sqrt t},\dfrac3{t^2}\right) \end{aligned}$ In conclusion, $f'(t)=\left(-\dfrac2{\sqrt t},\dfrac3{t^2}\right)$.